# Center of mass momentum of a photon and electron system

• e2m2a
In summary: So given an electron with a given energy/momentum and a photon with a given energy/momentum then after absorption there would be two unknowns, the electron energy and momentum. However there are three equations:...After absorption, there are three unknowns: the electron energy, momentum, and the photon's energy and momentum. However, there are two equations: the conservation of energy and momentum. After absorption, there are three unknowns: the electron energy, momentum, and the photon's energy and momentum. However, there are two equations: the conservation of energy and momentum.

#### e2m2a

Suppose there is a photon with momtum p=h/lambda moving in the positive x-direction. Suppose it collides with an electron at rest and is completely absorbed by the electron, and that after the collision, the electron moves to the right with the same momentum of the photon. This seems necessary because of the conservation of momentum. But how would we anyalzye this in terms of the momentum of the center of mass of the system? Can we even do this since the photon has no rest mass?

Your proposed process cannot occur. It would violate energy or momentum conservation.

Orodruin said:
Your proposed process cannot occur. It would violate energy or momentum conservation.
Then, what would happen?

e2m2a said:
Then, what would happen?
What would happen when? You have described an impossible process.

Ok. What about the compton effect, where a high energy x-ray photon collides with an electron? Isn't the momentum of the system conserved?

In the Compton effect, a photon scatters off of an electron. It is not absorbed. Both energy and momentum are conserved.

Last edited:
vanhees71
e2m2a said:
Suppose it collides with an electron at rest and is completely absorbed by the electron
This is not possible. You could have an atom absorb a photon, but not an electron. There is no way to conserve energy and momentum for the electron since it has no internal degrees of freedom.

Heikki Tuuri and vanhees71
During the absorption / emission of a photon the total momentum and energy of the system is conserved. Now the system contains both photon and electron.

GreatestPhysician99 said:
During the absorption / emission of a photon the total momentum and energy of the system is conserved. Now the system contains both photon and electron.
Momentum and energy are always conserved no matter what.

GreatestPhysician99 said:
Momentum and energy are always conserved no matter what.
Yes. This is precisely why an isolated electron cannot absorb a photon. An isolated electron cannot absorb a photon while conserving energy and momentum. Therefore the process is forbidden.

Dale said:
Yes. This is precisely why an isolated electron cannot absorb a photon. An isolated electron cannot absorb a photon while conserving energy and momentum. Therefore the process is forbidden.
Hmm why is it forbidden ? The photon's momentum and energy will be converted to electron's momentum and energy since the electron absorbes the photon.If it emmits it , the electron will lose momentum and energy...

GreatestPhysician99 said:
Hmm why is it forbidden ? The photon's momentum and energy will be converted to electron's momentum and energy since the electron absorbes the photon.If it emmits it , the electron will lose momentum and energy...
Having no internal structure, the only energy the electron can possesses is kinetic. If we adopt a frame of reference in which the electron winds up at rest, we have an electron arriving with non-zero kinetic energy and combining with a photon that has non-zero kinetic energy. The result is an electron at rest. This obviously violates conservation of energy.

Dale
jbriggs444 said:
Having no internal structure, the only energy the electron can possesses is kinetic. If we adopt a frame of reference in which the electron winds up at rest, we have an electron arriving with non-zero kinetic energy and combining with a photon that has non-zero kinetic energy. The result is an electron at rest. This obviously violates conservation of energy.
The electron will gain momentum (start moving ) according to the static frame of reference . So this is wrong.

GreatestPhysician99 said:
The electron will gain momentum (start moving ) according to the static frame of reference . So this is wrong.
There is no such thing as a "static frame of reference". All inertial frames of reference are equally valid, including the one in which the electron winds up at rest. Energy is required to be conserved in all of them. Momentum is required to be conserved in all of them.

And when it emmits a photon it will lose momentum (slow down) according to the static frame of reference

GreatestPhysician99 said:
And when it emmits a photon it will lose momentum (slow down) according to the static frame of reference
Again, this is simply incorrect.

jbriggs444 said:
There is no such thing as a "static frame of reference". All inertial frames of reference are equally valid, including the one in which the electron winds up at rest. Energy is required to be conserved in all of them. Momentum is required to be conserved in all of them.
It can be conserved as relativistic momentum or energy.

GreatestPhysician99 said:
The photon's momentum and energy will be converted to electron's momentum and energy since the electron absorbes the photon.
The math doesn’t work out. Remember that energy, mass, and momentum are related by ##m^2 c^2=E^2/c^2-p^2##. The mass of a photon is 0 and the mass of an electron is 511 keV/c.

So given an electron with a given energy/momentum and a photon with a given energy/momentum then after absorption there would be two unknowns, the electron energy and momentum. However there are three equations: the equation above, conservation of energy, and conservation of momentum. There is no solution to this set of three equations in two unknowns. Therefore absorption is forbidden.

anorlunda
It's easy to do this in the zero 3-momentum frame. In that frame, the electron's 4-momentum is ##(\gamma m_ec,\gamma m_ev,0,0)##. By the definition of a zero-momentum frame the x component of the photon's momentum must be equal and opposite to the electron's, and the t component must have the same absolute value because it's null. The photon's 4-momentum is thus ##(\gamma m_ev,-\gamma m_ev,0,0)## and the total 4-momentum is therefore ##(\gamma m_e(c+v),0,0,0)##.

If the photon is absorbed by the electron we are left with a single particle with velocity zero and mass ##m_e\gamma (c+v)/c##. This does not describe an electron. Therefore an electron cannot absorb a photon.

Ibix said:
It's easy to do this in the zero 3-momentum frame. In that frame, the electron's 4-momentum is ##(\gamma m_ec,\gamma m_ev,0,0)##. By the definition of a zero-momentum frame the x component of the photon's momentum must be equal and opposite to the electron's, and the t component must have the same absolute value because it's null. The photon's 4-momentum is thus ##(\gamma m_ev,-\gamma m_ev,0,0)## and the total 4-momentum is therefore ##(\gamma m_e(c+v),0,0,0)##.

If the photon is absorbed by the electron we are left with a single particle with velocity zero and mass ##m_e\gamma (c+v)/c##. This does not describe an electron. Therefore an electron cannot absorb a photon.
If one is anyway using 4-vectors, there is also a point in doing it without any reference to a frame. Calling the 4-momenta of the initial electron, final electron, and photon ##p_i##, ##p_f##, and ##k##, respectively, 4-momentum conservation reads ##p_i + k = p_f##. Squaring this we obtain (in units where ##c = 1##)
$$p_f^2 = m_e^2 = (p_i+k)^2 = p_i^2 + k^2 + 2p_i\cdot k = m_e^2 + 0 + 2p_i \cdot k > m_e^2,$$
since ##p_i \cdot k > 0## due to the electron 4-momentum being time-like and being a non-zero null vector. Thus, we reach the false inequality ##m_e^2 > m_e^2## (note that the inequality is strict), which means the proposed process cannot satisfy 4-momentum conservation.

Ibix